Problem 82
Question
Factor. If an expression is prime, so indicate. $$ -28 u^{3} v^{3}+26 u^{2} v^{4}-6 u v^{5} $$
Step-by-Step Solution
Verified Answer
\( uv^3(-14u + 6v)(2u + v) \).
1Step 1: Identify the Greatest Common Factor (GCF)
First, we look for the greatest common factor (GCF) of the terms in the expression \(-28u^3v^3 + 26u^2v^4 - 6uv^5\).Each term contains at least one \(u\) and three \(v\)'s, so we can factor out \(uv^3\) from all terms.
2Step 2: Factor Out the Greatest Common Factor
Factor out the GCF, \(uv^3\), from the expression:\[ uv^3(-28u^2 + 26uv - 6v^2) \].
3Step 3: Factor the Trinomial
Now, work on factoring the trinomial \(-28u^2 + 26uv - 6v^2\).To factor this, find two numbers that multiply to \(-28 \times -6 = 168\)and add to \(26\). The numbers \(-14\) and \(12\) satisfy this requirement:\( (-14+12=26) \).
4Step 4: Rewrite and Factor by Grouping
Rewrite the middle term \(26uv\) by breaking it into the two numbers found: \(-28u^2 - 14uv + 12uv - 6v^2\).Next, factor by grouping:1. From the first two terms: \(-28u^2 - 14uv = -14u(2u + v)\).2. From the last two terms: \(12uv - 6v^2 = 6v(2u + v)\).Our expression is now:\[-14u(2u + v) + 6v(2u + v) \].
5Step 5: Factor out the Common Binomial
There is a common factor of \(2u + v\) in both groups:\[ (-14u + 6v)(2u + v) \].Thus, the expression becomes:\[ uv^3(-14u + 6v)(2u + v) \].
6Step 6: Final Simplification
Double-check all factored terms to ensure the solution is simplified completely. The terms within and outside the parenthesis are already in simplest form.So, the final factored form is:\[ uv^3(-14u + 6v)(2u + v) \].
Key Concepts
Greatest Common FactorTrinomial FactoringFactor by Grouping
Greatest Common Factor
When we begin the process of factoring polynomials, the first step is often to identify the Greatest Common Factor (GCF). The GCF is the largest factor that all terms in an expression share. By factoring out the GCF, we simplify the expression, making it easier to work with in subsequent steps.
In the expression \(-28u^3v^3 + 26u^2v^4 - 6uv^5\), we look for the highest powers of variables and coefficients that are common in each term. Here, each term contains at least one \(u\) and three \(v\)s. By identifying this, we determine \(uv^3\) as the GCF. This simplifies the process of factoring because we can pull out \(uv^3\) from the whole expression, reducing it to \(uv^3(-28u^2 + 26uv - 6v^2)\).
Finding the GCF is a fundamental part of simplifying expressions and setting the groundwork for further factoring techniques.
In the expression \(-28u^3v^3 + 26u^2v^4 - 6uv^5\), we look for the highest powers of variables and coefficients that are common in each term. Here, each term contains at least one \(u\) and three \(v\)s. By identifying this, we determine \(uv^3\) as the GCF. This simplifies the process of factoring because we can pull out \(uv^3\) from the whole expression, reducing it to \(uv^3(-28u^2 + 26uv - 6v^2)\).
Finding the GCF is a fundamental part of simplifying expressions and setting the groundwork for further factoring techniques.
Trinomial Factoring
Trinomial factoring involves breaking down an expression with three terms into two binomial expressions. This step comes after identifying and factoring out any GCF.
Consider the trinomial \(-28u^2 + 26uv - 6v^2\). To factor it, we need two numbers that multiply to the product of the leading coefficient and the constant term. Here, this product is \(-28 imes -6 = 168\). The goal is to find two numbers that add up to the middle coefficient, 26. These numbers are -14 and 12, since \(-14 + 12 = 26\).
By using these numbers, we rewrite the middle term \(26uv\) in terms of these factors, setting the stage for factor by grouping: \(-28u^2 - 14uv + 12uv - 6v^2\). Proper trinomial factoring simplifies the expression and prepares it for factoring by grouping.
Consider the trinomial \(-28u^2 + 26uv - 6v^2\). To factor it, we need two numbers that multiply to the product of the leading coefficient and the constant term. Here, this product is \(-28 imes -6 = 168\). The goal is to find two numbers that add up to the middle coefficient, 26. These numbers are -14 and 12, since \(-14 + 12 = 26\).
By using these numbers, we rewrite the middle term \(26uv\) in terms of these factors, setting the stage for factor by grouping: \(-28u^2 - 14uv + 12uv - 6v^2\). Proper trinomial factoring simplifies the expression and prepares it for factoring by grouping.
Factor by Grouping
Factor by grouping is a useful method when dealing with expressions that are structured as a quadratic or more complex polynomial. After rewriting the trinomial using numbers determined in trinomial factoring, we group the terms strategically.
For the expression \(-28u^2 - 14uv + 12uv - 6v^2\), we group as follows:
This results in the expression: \((-14u + 6v)(2u + v)\). By completing this step carefully, it is possible to simplify even complex polynomials effectively, making the factor by grouping technique particularly valuable in higher-level algebra.
For the expression \(-28u^2 - 14uv + 12uv - 6v^2\), we group as follows:
- First Group: \(-28u^2 - 14uv\), which factors into \(-14u(2u + v)\)
- Second Group: \(12uv - 6v^2\), factoring into \(6v(2u + v)\)
This results in the expression: \((-14u + 6v)(2u + v)\). By completing this step carefully, it is possible to simplify even complex polynomials effectively, making the factor by grouping technique particularly valuable in higher-level algebra.